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Hunter-Best [27]
2 years ago
15

How many grams are in 6.8L of Oxygen gas (O2) at STP

Chemistry
1 answer:
zubka84 [21]2 years ago
4 0

Answer:

9.7 g

Explanation:

From the question,

Note: The molar volume of all gas at stp is 22.4 dm³ or 22.4 L

1 mol of  oxygen gas (O₂) at stp = 22.4 dm³

X mole of oxygen gas (O₂) at stp = 6.8 L

X = (1 mol×6.8 L)/22.4 L

X = 0.3036 mol.

But,

Number of mole (n) = mass (m)/molar mass (m')

n = m/m'

m = n×m'.................. Equation 2

Where n = 0.3036 mol, m' = 32 g/mol

Substitute into equation 2

m = 0.3036×32

m = 9.7 g

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Elements in the same group have the same number of valence electrons after their group number.

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You may find bellow the balanced chemical equations.

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Ionic equations:

3 Sr²⁺ (aq) + 6 NO₃⁻ (aq) + 6 K⁺ (aq) + 2 PO₄³⁻ (aq) →  Sr₃(PO₄)₂ (s) + 6 K⁺ (aq) + 6 NO₃⁻ (aq)

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To get the net ionic equation we remove the spectator ions:

3 Sr²⁺ (aq) + 2 PO₄³⁻ (aq) →  Sr₃(PO₄)₂ (s)

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brainly.com/question/7018960

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